Superposition Principle • When two or more wave disturbances occur at the same point, the total disturbance is just their WAVE INTERFERENCE algebraic sum:! AND NORMAL MODES y(x,t) = A1sin("1t–k1x) + A2sin("2t–k2x) Parts of Chapters 15 & 16 •! Adding disturbances is linear: if wave 1 and Young & Freedman, 11th ed. wave 2 satisfy the wave equation, their linear combination must also satisfy it •! This is the principle of linear superposition 05/30/2006 Oscillations & Waves 1 05/30/2006 Oscillations & Waves §15.6 2 1 2 Interference of Waves Boundary Conditions •! When 2 or more waves overlap in the same space • A disturbance reflected from a fixed end of a interference occurs. medium is inverted. This corresponds to two •! Constructive interference means the waves add up inverted waves passing in opposite directions. to produce increased amplitude (waves in-phase) • A disturbance reflected from a free end of a •! Destructive interference means the waves subtract medium is not inverted. This corresponds to two to produce reduced amplitude (waves out-of-phase) identical waves passing in opposite directions. ! For sinusoidal waves, “inversion” is equivalent to a phase change of radians (180°) Completely constructive if Completely destructive if ! 2 identical waves in-phase 2 identical waves ! rad out-of-phase 05/30/2006 Oscillations & Waves §16.6 3 05/30/2006 Oscillations & Waves §15.6 4 3 4 Standing Waves • What happens if two identical sinusoidal waves, travelling in opposite directions, are superimposed? • Alternating constructive & destructive interference • Node & Antinode positions don’t move 05/30/2006 Oscillations & Waves §15.6 05/30/2006 Oscillations & Waves 5 §15.7 6 5 6 Standing Waves (2) Standing Waves (3) • At x, string oscillates with angular frequency " • From superposition, and amplitude 2Asinkx y(x,t) = A[sin("t + kx) – sin("t – kx)] • All points between two adjacent nodes oscillate in • Using sin(a ± b) = sin a cos b ± cos a sin b phase y(x,t) = A[sin("t)cos(kx) + cos("t)sin(kx)] – A[sin("t)cos(kx) – cos("t)sin(kx)] • Then “standing waves” have the form y(x,t) = 2Asin(kx)cos("t) 05/30/2006 Oscillations & Waves §15.7 7 05/30/2006 Oscillations & Waves §15.7 8 7 8 Nodes, antinodes, and boundary Standing waves and travelling conditions waves • Places where sin(kx) = 0 are called nodes. At • A standing wave is a superposition of two these points kx = n!, or x = n#/2 travelling waves — one travelling to the (n = 0,1,2,…) right and the other to the left • Antinodes occur where sin(kx) = ±1. • Only certain standing waves are allowed on a string, in a vibrating air column, etc. There kx = n!/2, or x = n#/4 (n odd) • Other waves are allowed, but will not make • If end points are fixed they must be nodes; free steady standing waves (i.e. they will end points correspond to antinodes eventually cancel out; “transient”) • No net energy transport for standing waves 05/30/2006 Oscillations & Waves §15.7 9 05/30/2006 Oscillations & Waves §15.7 10 9 10 Example 15.7 Example 15.7 (2) A guitar string lies on the x-axis from x = 0 to L. A Then y(x,t) = 2A sin(kx) cos("t) sinusoidal wave with v = –143 m/s, A = 0.750 mm = (1.50mm) sin(19.3rad/m)x cos(2760rad/s)t and ƒ = 440 Hz is reflected from The node positions are at x = n#/2(n=0,1,2,…) x = 0 to form a standing wave. where # = v/ƒ = 143m.s–1/440Hz = 0.325m Find y(x,t), xn, & ASW, (vy)max, amax at antinodes. so x = 0, 0.163m, 0.325m, 0.488m, … As indicated above, ASW = 2A = 1.50mm Solution: First we need " and k. (vy)max = "ASW = 4.15 m/s " = 2!ƒ = 2! rad x 440 Hz = 2760 rad/s 2 4 2 amax = " ASW = 1.15x10 m/s = 1170g!! k = "/v = 2760 rad.s–1/143m.s–1 = 19.3 rad/m 05/30/2006 Oscillations & Waves 11 05/30/2006 Oscillations & Waves 12 11 12 Normal modes of a stretched Normal modes string • These special standing waves are called the normal modes of the string or air column etc. • Length of string must be #/2, #, 3#/2, etc: • In a normal mode, all oscillating particles of the ! ! L = n#/2, n = 1, 2, …! system move sinusoidally with the same • Conversely, the possible values of # are frequency ! ! #n = 2L/n, n = 1, 2, 3, … • In a normal mode, all oscillating particles of the • Each normal mode has a different frequency ($ is system are either exactly in phase or exactly ! wave speed) radians out of phase ! fn = n$/(2L) = nf1 , n = 1, 2, 3, … , • Normal mode frequencies are resonant frequencies05/30/2006 Oscillations & Waves §15.8 13 05/30/2006 Oscillations & Waves §15.8 14 13 14 Normal modes of a stretched string (2) 1 F f = 1 2L µ • Thus for a musical ! instrument, changing F, L, µ tunes the note 05/30/2006 Oscillations & Waves §15.8 15 15.